Mensuration Exercise 15.5 – Questions

A petrol tank is in the shape of a cylinder with hemisphere of same radius attached to both ends. If the total length of the tank is 6m and the radius is 1m, what is the capacity of the tank in litres.

A rocket is in the shape of a cylinder with a cone attached to one end and a hemisphere attached to the other. All of them are of the same radius of 1.5m. The total length of the rocket is 7m and height of the cup is 2m. Find the volume of the rocket.

A cup is in the form of a hemisphere surmounted by a cylinder. The height of the cylindrical portion is 8cm and the total height of the cup is 11.5cm. Find the TSA of the cup.

A storage tank consists of a circular cylinder with a hemisphere adjoined on either ends. The external diameter of the cylinder is 1.4 m and length is 8m, find the cost of painting it on the outside at the rate of Rs. 10 per m2.

A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is 16 cm and its height is 15 cm. Find the cost of painting the toy at 7 per Rs. 100 cm2.

A circus tent is cylindrical up-to a height of 3m and conical above it. If the diameter of the base is 105m and the slant height of the conical part is 53m, find the total cost of canvas used to make the tent if the cost of the canvas per m2 is Rs. 100.

Mensuration Exercise 15.5 – Solutions

A petrol tank is in the shape of a cylinder with hemisphere of same radius attached to both ends. If the total length of the tank is 6m and the radius is 1m, what is the capacity of the tank in litres.

Solution:

r = 1m

total length of the tank = 6m

h = 6 – (1 + 1) = 6 – 2 = 4 m

Volume of the tank = volume of the hemisphere + volume of the cylinder + volume of the hemisphere

= 2/3πr3 + πr2h + 2/3πr3

= 2/3 x 22/7­ x 13 + 22/7 x 12 x 4 + 2/3 x 22/7 x 13

= 16.76 m

Capacity of the water tank in litres = 16.76 x 1000 = 16761.9 l

A rocket is in the shape of a cylinder with a cone attached to one end and a hemisphere attached to the other. All of them are of the same radius of 1.5m. The total length of the rocket is 7m and height of the cup is 2m. Find the volume of the rocket.

Solution:

r = 1.5 m

Height of the cylinder, h = 7 – (1.5 + 2) = 7 – 3.5 = 3.5 m

height of the cone = 2m

Volume of the rocket = Volume of the cone + volume of the cylinder + volume of hemisphere

= 1/3πr2h+ πr2h + 2/3πr3

= 1/3 x 22/7 x (1.5)2 x 2 + 22/7 x (1.5)2 x 3.5 + 2/3 x 22/7 x (1.5)3

= 36.53 m3

A cup is in the form of a hemisphere surmounted by a cylinder. The height of the cylindrical portion is 8cm and the total height of the cup is 11.5cm. Find the TSA of the cup.

Solution:

total height of the cup = 11.5 cm

Height of the cylindrical portion = 8 cm

Height of the hemisphere = 11.5 – 8 = 3.5 cm

r = 3.5 cm

TSA of the cup = TSA of cylinder + TSA of hemisphere

= 2πrh + 2πr2

= 2 x 22/7 x 3.5 x 8 + 2 x 22/7 x 3.52

= 253 cm2

A storage tank consists of a circular cylinder with a hemisphere adjoined on either ends. The external diameter of the cylinder is 1.4 m and length is 8m, find the cost of painting it on the outside at the rate of Rs. 10 per m2.

Solution:

External diameter = 1.4 m

Radius = d/2 = 1.4/2 = 0.7 m

length = 8m

length of the hemisphere = 8 – (0.7+0.7) = 6.6 m

Total surface area of the storage tank = TSA of the cylindrical part + 2x TSA of hemispherical part

A circus tent is cylindrical up-to a height of 3m and conical above it. If the diameter of the base is 105m and the slant height of the conical part is 53m, find the total cost of canvas used to make the tent if the cost of the canvas per m2 is Rs. 100.

Solution:

d = 105 m

r = d/2 = 105/2 = 52.5 m

h = 3m

l = 53 m

l2 = r2 + h2

532 = 52.52 + h2

532 – 52.52 = h2

52.75 = h2

h = 7.26 m

Total canvas used = CSA of cylindrical part + CSA of conical part

= 2πrh + πrl

= 2 x 22/7 x 52.5 x 3 + 22/7 x 52.5 x 53

= 9735 m2

The total cost of canvas used to make the tent if the cost of the canvas per m2 is Rs. 100 = 9735 x 100 = Rs. 97,35,000

After studying the chapter Ratio and Proportion and their general form; to understand and differentiate between different types of proportion; to acquire skills of writing proportion; to solve problems on time and work involving proportions; to apply proportion in day to day life situations.

2.4.1 Introduction to Ratio and Proportion

In a ratio a : b, the first term a is called the antecedent and the second term b is called the consequent. Ratio is an abstract quantity and has no unit. Ratio tells how many times the first term is there in the second term.

Example 1: In the adjacent figure, find the ratio of the shortest side of the triangle to the longest side.

Solution:

We see that the shortest side is of length 5 cm and the longest side is of length 13 cm. Hence the ratio is 5:13

Example : Suppose the ratio of boys to girls in a school of 720 students is 7 : 5. How many more girls should be admitted to make the ratio 1 : 1?

Solution:

For every 7 boys there are 5 girls. Thus out of 12 students, 7 are boys and 5 are girls. Hence the number of boys is 7/12 x 720 = 420.

The number of girls is 720 – 420 = 300. Now we want the ratio of boys to girls to be 1:1. This means the number of boys and girls must be same. Since the deficiency of girls is 420 – 300 = 120, the school must admit 120 girls to make the ratio 1:1.

Example 5: Consider the ratio 12:5 If this ratio has to be reduced by 20% which common number should be added to both the numerator and denominator?

Solution:

Consider 12/5. This has to be reduced by 20%. This means we have to consider 80% of this number. Thus, we must get,

12/5 x 80/100 = 48/25

We have to find a such that,

12+a/5+a = 48/25

Cross multiplying

25(12 + a) = 48(5 + a)

48a – 25a = (25 x 12) – (48 x 5) =

23a = 60

a = 60/23

If we add 60/23 to both terms of 12 : 5 we get a ratio which is 20% less than the original ratio.

Ratio and Proportion – Exercise 2.4.1

1.Write each of these ratios in the simplest form.

(i) 2:6

(ii)24:4

(iii) 14:21

(iv) 20: 100

(v) 18:24

(vi) 22:77

Solution:

(i) 2:6 = 1:3 (dividing both by 2)

(ii) 24:4 = 6:1 (dividing both by 2)

(iii) 14:21 = 2:3 (dividing both by 7)

(iv) 20:100 = 1:5 (dividing both by 2)

(v) 18:24 = 3:4 (dividing both by 6)

(vi) 22:77 = 2:7 (dividing both by 11)

2. A shop-keeper mixes 600 ml of orange juice with 900 ml of apple juice to make a fruit drink. Write the ratio of orange juice to apple juice in the fruit drink in its simplest form.

Solution:

Ratio of volumes of

Orange juice and apple juice O:A

= 600:900

= 6:9

= 2:3

3. a builder mixes 10 shovels of cement with 25 shovels of sand. Write the ratio of cement to sand.

Solution:

Ratio of cement to sand = 10 shovels :25 shovels

4.In a school there are 850 pupils and 40 teachers. Write the ratio of teachers to pupils.

Solution:

Number of teachers : Number of pupils

= 40 : 850 = 4:85

5. On a map, a distance of 5cm represent an actual distance of 15km. Write the ratio of the scale of the map.

Solution:

Let x be the number to be added them

(49 + x) = (68 + x) = 3:4

4(49+X) = (68 + X)3

196+4X = 204 + 3X

4X – 3X = 204 – 196

x = 8

2.4.2 Proportion – Ratio and Proportion

Ratio and Proportion – Exercise 2.4.2

1. In the adjacent figure, two triangles are similar find the length of the missing side

Solution:

Let the triangles be ABC and PQR

BC/QR = AC/PR

5/X = 13/39

13X = 5 X 39

X = 5X39/13= 5 X3 = 15

What number is to 12 is 5 is 30?

Solution

Let x be the number

x:12 :: 5 : 30

30x = 12×5

x = 12×5/30 = 2

Solve the following properties:

(i). x : 5 = 3 : 6

(ii) 4 : y = 16 : 20

(iii) 2 : 3 = y : 9

(iv) 13 : 2 = 6.5 : x

(v) 2 : π = x : 22/7

Solution:

(i). x : 5 = 3 : 6

6x = 5 x 3

6x = 15

x = 15/6

(ii) 4 : y = 16 : 20

4×20 = 16y

y = 4×20/16

y = 5

(iii) 2 : 3 = y : 9

2×9 = 3y

y = 2×9/3 = 2×3 = 6

(iv) 13 : 2 = 6.5 : x

13x = 2 x 6.5

13x = 13

x = 13/13 = 1

(v) 2 : π = x : 22/7

2x22/7 = πx

x = (2x22/7) /π =(2x22/7) /(22/7)

x = 2

find the mean proportion to :

(i) 8, 16

(ii) 0.3, 2.7

(ii)162/3 , 6

(iv) 1.25, 0.45

Solution:

(i) 8, 16

Let x be the mean proportion to 8 and 16

Then 8/x = x/16

x2 = 8 x 16 = 128

x = √128 = √(64×2) = 8√2

(ii) 0.3, 2.7

Let x be the mean proportion to 0.3 and 2.7

Then 0.3/x = x/2.7

x2 = 0.3 x 2.7 = 0.81

x = √(0.81) = 0.9

(ii)162/3 , 6

Let x be the mean proportion to 162/3 and 6

Then (162/3) /x = x/6

x2 = 162/3 x 6 = 50/3 x 6 = 100

x = √100 = 10

(iv) 1.25, 0.45

Let x be the mean proportion to 1.25 and 0.45

Then 1.25/x = x/0.45

x2 = 1.25 x 0.45

x = √(1.25 x 0.45) = √(1.25 x 0.45)x√(100×100)/ √(100×100)

= √(125×45)/√(100×100) = √(25x5x5x9)/√(100×100) = 5x5x3/10×10= 3/4

Find the fourth proportion for the following:

(i) 2.8, 14, 3.5

(ii) 31/3, 12/3, 21/2

(iii)15/7, 23/4, 33/5

Solution:

(i) 2.8, 14, 3.5

Let x be the fourth proportion

Then, 2.8 : 14 :: 3.5 : x

2.8x = 14×3.5

x = 14×3.5/2.8 = 17.5

(ii) 31/3, 12/3, 21/2

Let x be the fourth proportion

Then, 31/3: 12/3 :: 21/2: x

10/3 :5/3 : : 5/2 : x

10/3x = 5/3 x 5/2

10/3x = 25/6

x = 25/6 x 3/10 = 75/60 = 15/12 = 5/4

(iii)15/7, 23/14, 33/5

Let x be the fourth proportion

Then, 15/7: 23/14:: 33/5: x

12/7 :31/14 : : 18/5 : x

12/7 x = 31/14 x 18/5

12/7 x = 31×18/14×5

x = 31×18/14×5 x 7/12 = 31×3/5×4 = 93/20 = 413/20

Find the third proportion to:

(i) 12, 16

(ii) 4.5, 6

(iii) 51/2 , 161/2

Solution:

(i) 12, 16

Let x be the third proportion

Then 16:12 :: x : 16

12x = 16 x16

x = 16×16/12 = 64/3 = 211/3

(ii) 4.5, 6

Let x be the third proportion

Then 6:4.5 :: x : 6

4.5x = 6 x6

x = 6×6/4.5 = 36/4.5 = 360/45 = 8

(iii) 51/2 , 161/2

Let x be the third proportion

Then 161/2: 51/2:: x : 161/2

11/2 x = 33/2 x 33/2

x = 33/2 x 33/2 x 2/11= 33×3/2 = 99/2 = 491/2

In a map 1/4 cm represents 25km, if two cities are 21/2c apart on the map, what is the actual distance between them?

Solution:

Let 21/2 cm represemts x km

1/4cm: 25km :: 21/2cm : x km

1/4 x x = 25 x 21/2

x/4 = 25 x5 /2

x = 25×5/2 x 4 = 25x5x2 = 250km

Suppose 30 out of 500 components for a computer were found defective. At this rate how many defective components would he found in 1600 components?

Solution:

Number of defective components in 500 components = 30

Let x be the number of defective components in 1600 components

then 30:500 :: x :1600

30×1600 = 500x

x = 30×1600/500 =96

2.4.3 Time and Work – Ratio and Proportion

Ratio and Proportion – Exercise 2.4.3

Suppose A and B together can do a job in 12 days, while B alone can finish a job in 24 days. In how many days can A alone finish the work?

Solution:

Number of days in which A and B together can finish the work = 12 days

Number of days in which B alone can finish the work = 30

1/T = 1/m + 1/n

1/12 = 1/m + 1/30

1/m = 1/30 – 1/12 = 5-2/60 = 3/60 = 1/20

A can finish the work in 20 days.

Suppose A is twice as good a workman as B and together they can finish a job in 24 days. How many days A alone takes to finish the job?

Solution:

A is twice as good a workman as B

i.e if B can finish a work in t days A can finish it in 1/2 days

1/T = 1/m + 1/n

1/24 = 1/t/2 + 1/t = 2/t + 1/t = 3/t

1/24 = 3/t

t = 24 x 3 = 72

i.e, B takes 72days to finish the job

A takes 72/2 = 36 days to finish it

Suppose B is 60% more efficient them A. if A can finish a job in 15 days how many days B needs to finish the same job?

Solution:

A can finish a work in 15 days.

Work done A in 1 day = 1/15

B is 60% more efficient

Work done by B in 1 day

1/15 + 1/15 x 60/100

= 1/15 (1 + 60/100)

= 1/15 ( 8/5)

= 8/75

Number of days in which B alone can finish the work = 1/(8/75) = 75/8 = 93/8 days

Suppose A can do a piece of work in 14 days while B can do it in 21 days. They begin together and worked at it for 6 days. Then A fell ill B had to complete the work alone. In how many days was the work completed?

Solution:

M = 14 days

N = 21 days

Part of work done in 6 days

= (1/14 + 1/21)6

= 6(3+2/42) = 5×6/42 = 5/7

Remaining part of the work = 1-5/7 = 2/7

Days taken by B to finish

2/7 part of the work = (2/7)/(1/21) = 2/7 x 21/7 = 6 days

Total number of days in which the work is completed = 6+6 = 12 days

Suppose A takes twice as much time as B and thrice as much time as C to complete a work. If all of them work together they can finish the work in 2 days. How much time B and C working together will take to finish it?

Solution:

If A alone takes to t1 days to do the work , B finishes it in t1/2 and C is t1/3 days

1/T = 1/t1 +1/t2 + 1/t3

= 1/t1 +1/(t1/2) + 1/(t1/3)

= 1/t1 +2/t1 + 3/t1

= 6/t1

1/T = 1/2

1/2 = 6/T1

i.e. t1 = 12 dyas

B takes 12/2 = 6days

C takes 12/3 = 4 days

Part of work done by B

In one day = 1/6

Part of work done by C in one day = 1/4

If B and C together takes t days to finish the work 1/T = 1/6 + 1/4 = 2+3/12 = 5/12

How many perfect cubes are there from 1 to 500? How many are perfect square among cubes?

Find the cubes of 10, 30 , 100, 1000. What can you say about the zeros at the end?

Solution:

103 = 1000

303 = 27000

1003 = 1000000

10003 = 10000000000

The number of zero of a cube are 3 times, the no. of zero of numbers

What are the digits in the unit’s place of the cubes 1, 2, 3, 4, 5, 6, 7, 8 , 9, 10? Is it possible to say that a number is not a perfect cube by looking at the digit in unit’s place of the given number, just like you did for squares?

Solution:

a

a3

digit in the units place

1

1

1

2

8

8

3

27

7

4

64

4

5

125

5

6

216

6

7

343

3

8

512

2

9

729

9

10

1000

0

No, its not possible to tell a number is not a perfect cube by looking at the digit in unit’s place of the given number.

Examples to find perfect squares near to a given number

Example 5: If the area of a square is 90 cm2, what is its side-length rounded to the nearest integer?

Solution:

Since, area A = a2, we have a2 = 90.

But 81 < 90 < 100 and 81 is the nearer to 90 than 100. Hence, the nearest integer to √90 is √81 = 9.

Example 6: A square piece of land has area 112m2. What is the closest integer which approximates the perimeter of the land?

Solution:

If a is the side length of a square, its perimeter is 4a. We know that a2 = 112 m2.

Hence, (4a)2 = 16a2 = 16 x 112 = 1792

But 422 = 1764 < 1792 < 1849 = 432 .

!764 is nearer to 1792 than 1849. Therefore, the integer approximation for √1792 is 42. The approximate value of perimeter is 42cm.

Perfect squares near to a given number – Exercise 1.2.6

Find the nearest integer to the square root of the following numbers:

i) 232

Solution:

We have, 152 = 225 and 162 = 256.

We know, 225 < 232 < 256

152 < 232 < 162

∴ Square root of 232 is nearest to 15

ii) 600

Solution:

We have, 242 = 576 and 252 = 625.

We know, 576 < 600 < 625

242 < 600 < 252

∴ Square root of 600 s nearest to 24.

iii) 728

Solution:

We have, 262 = 676 and 272 = 729

676 < 728 < 729

262 < 728 < 272

∴ Square root of 728 is nearest to 27.

iv) 824

Solution:

We have, 282 = 784 and 292 = 841

784 < 824 < 841

282 < 824 < 292

∴Square root of 824 is nearest to 29

v) 1729

Solution:

We have, 412 = 1681 and 422 = 1764

1681 < 1729 < 1764

412 < 1729 < 422

∴Square root of 1729 is nearest to 42

A piece of land is in the shape of a square and its area is 1000m2 . This has to be fenced using barbed wire. The barbed wire is available only in integral lengths. What is the minimum length of the barbed wire that has to be bought for this purpose.

Solution:

Area of the land = 1000m2

Area = a2 = 1000m2

Perimeter = 4a

Squaring , (Perimeter)2 = ( 4a)2 = 16a2 = 16 x 1000

(perimeter)2 = 16,000

We have, square of the perimeter i.e., 16000. now we have to find the square root of 16000. As 16000 does not have a perfect square root, let us find the square root nearest to square root of 16000.

We know, 1262 = 15876 and 1272 = 16129.

Then, 15876 < 16000 < 16129

1262 < 16000< 1272

∴ Nearest number is 126 But this is not enough to cover the hard.

∴Length of barbed wire required = 127 m.

A student was asked to find √961. He read it wrongly and found √691 to the nearest integer. How much small was his number forms the correct answer?

Solution:

We know, √ 961 = 31

It is given that student read it wrongly and found the result for √691. So, we have to find out the difference between √961 and √691.

Example 3: Find the smallest positive integer with which one has to divide 336 to get a perfect square.

Solution:

We observe that 336 = 2 x 2 x 2 x 2 x 3 x 7. Here both 3 and 7 occur only once. Hence we have to remove them to get a perfect square.

We divide 336 by 3 and 7

42.

The required least number is 21.

Square root of a perfect square by factorization – Exercise 1.2.5

Find the square roots of the following numbers by factorization:

i) 196

Solution:

196 = 2 x 2 x 7 x 7

= (2 x 7) x (2 x 7)

= 14 x 14

196 = 142

ii) 256

Solution:

256 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

= (2 x 2 x 2 x 2) x (2 x 2 x 2 x 2)

= 16 x 16

= 162

iii) 10404

Solution:

10404 = 2 x 2 x 3 x 3 x 17 x 17

= (2 x 3 x 17) x (2 x 3 x 17)

= (102) x (102)

= 1022

iv) 1156

Solution:

1156 = 2 x 2 x 17 x 17

= (2 x 17) x (2 x 17)

= 34 x 34

= 342

v) 13225

Solution:

13225 = 5 x 5 x 23 x 23

= (5 x 23) x (5 x 23)

= 115 x 115

= 1152

2. Simplify:

i) √100 + √36

Solution:

√100 + √36 = 10 + 6

= 16

ii) √(1360 + 9)

Solution:

√(1360 + 9) = √1369 = √(37 x 37) = 37

iii) √2704 + √144 + √289

Solution:

√2704 + √144 + √289 = √(52 x 52) + √(12 x 12) + √(17 x 17)

= 52 + 12 + 17

= 81

iv) √225 – √25

Solution:

√225 – √25 = √(15 x 15) – √(5 x 5)

= 15 – 5

= 10.

V) √1764 – √1444

Solution:

√1764 – √1444 = √(42 x 42) – √(38 x 38)

= 42 – 38

= 4

vi) √169 x √361

√169 x √361 = √(13 x 13) x √(19 x 19)

= 13 x 19

= 247

A square yard has area 1764m2. From a corner of this yard, other square part of area 784m2 is taken out for public utility. The remaining portion is divided in to 5 equal parts. What is the perimeter of each of these equal parts?

Solution:

Area of square = a2

Area of given square yard = 1764 m2

Area used for public utility = 784 m2

∴ Area of remaining portion = 1764 – 784 = 980 m2

If the area is divided into 5 parts

Then, the area of each Square = 980/5 = 196 m2

Length of each side = √ 196, a = 14m

Perimeter of square = 4a = 4 x 14 = 56m

4. Find the smallest positive integer with which one has to multiply each of the following numbers to get a perfect square:

i) 847

Solution:

847 = 11 x 11 x 7

Here 7 occur only once. Hence we have to multiply by 7 them to get a perfect square.

ii) 450

Solution:

450 = 5 x 5 x 3 x 3 x 2

Here 2 occur only once. So we have multiply by 2 to get a perfect square.

iii) 1445

Solution:

1445 = 17 x 17 x 5

Here 5 occur only once. So we have multiply by 5 to get a perfect square.

iv) 1352

Solution:

1352 = 2 x 2 x 13 x 13 x 2

Here 2 occur only three times. So we have multiply by 2 to get a perfect square.

5. Find the largest perfect square factor of each of the following numbers:

i) 48

Solution:

48 = 2 x 2 x 2 x 2 x 3 = (2 x 2) x (2 x 2) x 3

= 4 x 4 x 3 = 16 x 3 = 42 + 3

Therefore, 16 is the largest perfect square factor of 48.

ii) 11280

Solution:

11280 = 2 x 2 x 2 x 2 x 3 x 47

Therefore, 16 is the largest perfect square factor of 11280

iii) 729

Solution:

729 = 3 x 3 x 3 x 3 x 3 x 3

= (27) x (27)

= 272

Therefore, 729 is the largest perfect square factor of 729.

iv) 1352

Solution:

1352 = 2 x 2 x 2 x 13 x 13

= (2 x 13) x (2 x 13) x 2

= 26 x 26 x 2

Therefore, 676 is the largest perfect square of 1352.

Find a proper positive factor of 48 and a proper positive multiple of 48 which add up to a perfect square. Can you prove that there are infinitely many such pairs?

Observe that, 1 = 1, 4 = 2 x 2, 9 = 3 x 3, 100 = 10 x 10. If a is an integer and b = a x a, we say b is a perfect square. Hence 1, 4, 9, 16, 25 are all perfect squares. Since 0 = 0 x 0, we see that 0 is a perfect square. If a is an integer, we denote a x a = a2. We read it as square of a or simply a square. Thus 36 = 62 and 81 = 92. Thus a perfect square is of the m2, where m is an integer.

For example, 4 = 2 x 2 and 4 = (-2) x (-2); in the second representation, we again have equal integers, but negative this time.

Thus a perfect square is either equal to 0 or must be a positive integer. It can be a negative integer.

Look at the following table

a

1

2

3

8

-7

-12

20

-15

a2

1

4

9

64

49

144

400

225

We see that squares of 2, 8, -12, 20 are even numbers and the squares of 1, 3, -7, -15 are odd numbers.

Statement 1: The Square of an even integer is even and the square of an odd integer is odd.

Consider first 10 perfect squares,

12

22

32

42

52

62

72

82

92

102

1

4

9

16

25

36

49

64

81

100

If we observe that the units place in these squares are 1, 4, 9, 6, 5, 6, 9, 4, 1 and 0 in that order. Thus only the digits which can occupy digit’s place in perfect squares are 1, 4, 5, 6 and 9.

Statement 2: A perfect square always ends in one of the digits 0, 1, 4, 5, 6 and 9. If the last digit of a number is 2, 3, 7 or 8, it cannot be a perfect square.

STATEMENT 7: consider the number N = 1000….01, where zeros appear k times. (For example, for k = 6, you get N = 10000001; there are 6 zeros in the middle.) Then N2 = 1000…02000…01, where the number of zeros on both the sides of 2 is k.

Ex:

112 = 121

1012 = 10201

10012 = 1002001

100012 = 100020001 and so on.

STATEMENT 8: The sum of nth and (n+1)th triangular number is (n+1)2.

Ex:

The dots are now arranged in shapes. Now count the number of dots in each triangle. (Single dot is considered as the generate triangle.) They are 1, 3, 6, 10, 15, 21, 28, 36 and so on. These are called triangular numbers.

For nth triangular number, we form a triangle of dots with n- rows and each row contains as many points as index of that row. If you want find the 8th triangular number, the number of points in the 8th triangle is

1.2.4 Methods of squaring a number – Perfect Squares

Many times it is easy to find the square of a number without actually multiplying the number to itself.

Consider 42 = 40 + 2

Thus,

422 = (40+2)2

= 402 + 2(40)(2) + 22

Now it is easy to recognise 402 = 1600; 2x40x2 = 160; 22 = 4.

Therefore, 422 = 1600 + 160 + 4 = 1764.

Example: Find 892

Solution: 892 = (80 + 9)2

= 802 + 2 x 80 x 9 + 92

= 6400 + 1440 + 81

= 7921

Perfect Squares – Exercise 1.2.4

Find the squares of:

i) 31

Solution:

31 = 30 + 1

312 = (30 + 1)2

= 302 + 2 x 30 x 1 +1

= 900 + 60 + 1

= 961

ii) 72

Solution:

72 = 70 + 2

722 = (70 + 2)2

= 702 + 2 x 70 x 2 + 22

= 4900 + 280 + 4

= 5184

iii) 37

Solution:

37 = 30 + 7

372 = (30 + 7)2

= 302 + 2 x 30 x 7 + 72

= 900 + 420 + 49

= 1369

iii) 166

Solution:

166 = 160 + 6

1662 = (160 + 6)2

= 1602 + 2 x 160 x 6 + 62

= 25600 + 1920 + 36

= 27556

2. Find the squares of:

i) 85

Solution:

852 = (80 + 5)2

= 802 + 2 x 80 x 5 + 52

= 1600 + 800 + 25

=2425

ii) 115

Solution:

1152 = (100 + 15)2

= 1002 + 2 x 100 x 15 + 152

= 10000 + 3000 + 225

= 13225

iii) 165

1652 = (160 + 5)2

= 1602 + 2 x 160 x 5 + 52

= 25600 + 1600 + 25

= 27225

Find the squares of 1468 by writing this as 1465+3

14682 = (1465 + 3)2

= (1465)2 + 2 x 1465 x 3 + 32

= 2146225 + 8790 + 9

= 2155024

1.2.5 Square roots – Perfect Squares

Consider the following perfect squares:

1 = 12, 4 = 22, 9 = 32, 16 = 42, 49 = 72, 196 = 142.

In each case the number is obtained by the product of two equal numbers. Here we say 1 is the square root of 1; 2 is the square root of 4; Hence 3 is the square root of 9 and so on.

Suppose N is a natural number such that N= M2. The number M is called a square root of N.

We have seen earlier m2 = m x m = (-m) x (-m). Thus m2 has 2 roots m and –m. For example, 16 = 42 = (-4)2, thus both 4 and -4 are the roots of 16. Mathematically both 4 and -4 are accepted as the square root of 16. Therefore, whenever the word square root is used, it is always meant to be the positive square root. The square root on n is denoted by √N.

If a number ends with any of the digits 0, 2, 4, 6, or 8, you immediately say the number is divisible by 2. Why? We write any such number a as a = 10k + r, where r is the remainder when divided by 10. Hence r is one of the numbers 0, 2, 4, 6, 8. We know see that 10 is divisible by 2 and r is also divisible by 2. We conclude that 2 divides a.

Divisibility tests – Playing with Numbers

Divisibility by 4 [Divisibility tests]

If a number is divisible by 4, it has to be divisible by 2. Hence the digit in the units place must be one of 0, 2, 4, 6, and 8. But look at the following numbers: 10, 22, 34, 46, and 58. We see that last digit in each of these numbers is as required, yet none of them is divisible by 4. Thus, we can conclude that it is not possible to decide the divisibility on just reading the last digit. Perhaps, the last two digits may help.

If a number has two digits, we may decide the divisibility by actually dividing it by 4. All we need is to remember the multiplication table for 4. Suppose the given number is large, say it has more than 2 digits. Consider the numbers, for example, 112 and 122. We see that 112 is divisible by 4. But 112 = 100 + 22; here 100 is divisible by 4 but 22 is not. Hence 122 is not divisible by 4.

We invoke the following fundamental principle on divisibility:

If a and b are integers which are divisible by an integer m ≠ 0, then m divides a+b, a-b and ab.

Now, let us see how does this help us to decide the divisibility of a large number by 4. Suppose we have number a with more than 2 digits. Divide this number by 100 to get a quotient q and remainder r; a = 100q + r, where 0 ≤ r < 100. Since 4 divides 100, you will immediately see that a is divisible by a 4 if and only if r is divisible by 4. But r is the number formed by the last two digits of a. Thus we may arrive at the following test:

STATEMENT 2: A number (having more than 2 digit ) is divisible by 4 if and only if the 2 digit number formed by the last two digits of a is divisible by 4.

Example 4: Check whether 12456 is divisible by 4.

Solution:

Here, the number formed by the last two digits is 56. This is divisible by 4 and hence so is 12456.

Example 5: Is the number 12345678 divisible by 4?

Solution:

The number formed by the last 2 digits is 78, which is not divisible by 4. Hence the given number is not divisible by 4.

Divisibility by 3 and 9 [Divisibility tests]

Consider the numbers 2, 23, 234, 2345, 23456, 234567. We observe that among these 6 numbers, only 234 and 234567 are divisible by 3. Here, we cannot think of the number formed by the last 2 digits or for those matter even three digits. Note that 3 divides 234, but it does not divide 34. Similarly, 3 divides 456 but it does not divide 23456.

STATEMENT 3: An integer a is divisible by 3 if and only if the sum of digits of a is divisible by 3. An integer b is divisible by 9 if and only if the sum of digits of b is divisible by 9.

Example 6: Check whether the number 12345321 is divisible by 3. Is it divisible by 9?

Solution:

The sum of digits is 1+2+3+4+5+3+2+1 = 21. Hence the number is divisible by 3, but not by 9. In fact 12345321 = (9 x 1371702) + 3.

Example 7: Is 444445 divisible by 3?

Solution:

The sum of digits is 25, which is not divisible by 3. Hence 444445 is not divisible by 3. Here the remainder is 1.

Divisibility by 5 and 10 [Divisibility tests]

Statement 4: An integer a is divisible by 5 if and only if it ends with 5 or 0. A number is divisible by 10 if and only if it ends with 0.

Note that 15 = 3 x 5. Here again the given number 12345 must be divisible by both 3 and 5.

The sum of the digits is 1+2+3+4+5 = 15, which is divisible by 3 and the given number 12345 ends with 5, so it’s also divisible by 5. Therefore, 12345 is divisible by 15.

Example 10: How many numbers from 201 to 250 are divisible by 5, but not by 3?

Solution:

The numbers which are divisible by 5 from 201 to 250 are 205, 210, 215, 220, 225, 230, 235, 240, 245, 250. Now let us compute the digital sum of these numbers:

205 ——-2+0+5 = 7, not divisible by 3

210 ——– 2+1 = 3, divisible by 3

215 ———2+1+5 = 8, not divisible by 3

220———–2+2+0 = 4, not divisible by 3

225———-2+2+5 = 9, divisible by 3

230———-2+3+0 = 5, not divisible by 3

235———-2+3+5 = 10, not divisible by 3

240———2+4+0 = 6, divisible by 3

245———-2+4+5 = 11, not divisible by 3

250———–2+5+0 = 7, divisible by 3

Divisible by 11 [Divisibility tests]

STATEMENT 5:

Given number n in decimal form, put alternatively – and + signs between the digits and compute them sum. The number is divisible by 11 if and only if this sum is divisible by 11. Thus a number is divisible by 11 and only if the difference between the sum of the digits in odd places and the sum of digits in even places is divisible by 11.

Example 11: Is the number 23456 divisible by 11?

Solution:

Observe that 2-3+4-5+6 = 4 and hence not divisible by 11. The test indicates that 23456 is not divisible by 11.

A palindrome is a number which leads the same from left to right or right to left. Thus a palindrome is a number n such that by reversing the digits of n, you get back n. For example, 232 is a 3 digit palindrome; 5445 is a 4 digit palindrome.

Example 12: Find all 3 digit palindromes which are divisible by 11.

Solution:

A 3 digit palindrome must be of the form aba, where a≠0 and b are digits. This divisible by 11 if and only if a-b+a = 2a-b is divisible by 11.

For a = 6, b = 1, we see that 2a-b = 12-1 = 11 and hence divisible by for which 2a-b is divisible by 11. We get four more numbers 616, 737, 858 and 979.

Thus required numbers are 121, 242, 363, 484, 616, 737, 858 979.

Example 13: Prove that 12456 is divisible by 36 without actually dividing it.

Solution:

First notice that 36 = 4 x 9. So it is enough if we prove 12456 is divisible by 4 and 9 both. The last 2 didgits of the given number 12456 is 56 which is divisible by 4 and it is left to show that, the given number 12456 is divisible by 9. Let us find the sum of digits of the given number 12456, i.e., 1+2+4+5+6 = 18, which is divisible by 9. Thus, the number 12456 is divisible by 36.

Divisibility tests – Playing with Numbers – Exercise 1.1.6

How many numbers from 1001 to 2000 are divisible by 4?

Solution:

We know that we have 25 numbers from 1001 to 1100. So we have 25X10 times = 250 numbers from 1001 to 2000 are divisible by 4.

Suppose a 3digit number abc is divisible by 3. prove that abc+bca+cab is divisible by 9.

Solution:

Let 3 digit number abc be 123, which is divisible by 3. Niw we have to prove, abc+bca+cab i.e., 123+231+312 is divisible by 9.

Thus, the cost of laying tiles on its floor and its four walls at the rate of Rs. 100/m²

= 50.88 x 100

= 50.88 Rs.

5. A closed box is 40cm long, 50cm wide and 60cm deep. Find the area of the foil needed for covering it.

Solution:

Total surface area of cuboid = 2(lb +bh + lh)

We have, l = 40 cm; b = 50 cm; h = 60 cm

Therefore,

Total surface area of cuboid = 2[(40 x 50) +(50 x 60) + (60 x 40)]

= 2[ 2000 + 3000 + 2400]

= 2[ 7400]

= 14800 cm²

6. The total surface area of a cube is 384cm². Calculate the side of the cube.

Solution:

Total surface area of cube = 6(l²)

384 = 6(l²)

l² = 384/6 = 64 cm²

l = √64 = 8m

Thus, The side of the cube is 8m.

7. The L.S.A of a cube is 64m². Calculate the side of the cube.

Solution:

Lateral surface area of cube = 4(l²)

64 = 4(l²)

l² = 64/4 = 16

Therefore, l = 4m

Thus, the side of the cube is 4l.

8.Find the cost of whitewashing the four walls of a cuboidal room of side 4m at the rate of Rs. 20/m²

Solution:

Lateral surface area of the cuboidal room = 4(l²)

= 4 ( 4²)

= 4 ( 16)

= 64 m²

The cost of whitewashing the cuboidal room at the rate of Rs. 20/m² = 64 x 20 = Rs. 1280

9. A cubical box has edge 10 cm and another cuboidal box is 12.5cm long, 10cm wide and 8cm high.

(i) Which box has smaller total surface total surface area?

(ii) If each edge of the cube is doubled, how many times will its T.S.A increase?

Solution:

(i)

Cube:

Total surface area = 6l² = 6 x 10² = 6 x 100 = 600m²

Cuboid:

Total surface area = 2(lb + bh + lh)

= 2[(12.5 x 10) + (10 x 8) + (8 x 12.5)]

= 2 [ 125 + 80 + 100]

= 610 m²

Thus, Cube has a smaller total surface area.

(ii)

If each edge of the cube is doubled,

Then,

Cube:

Total surface area = 6l² = 6 x 20² = 6 x 400 = 2400m²

Cuboid:

Total surface area = 2(lb + bh + lh)

= 2[(25 x 20) + (20 x 16) + (16 x 25)]

= 2 [ 500 + 320 + 400]

= 2[1220]

= 2440 m²

Thus,

If each edge of the cube is doubled, then its T.S.A will increase 4 times.

Volume of cubes and cuboids – Mensuration

Volume of cube = l³ , where l = length

Volume of cuboid = area of base x height;

where area of base = l x b, here l = length, b = breadth

Mensuration – Exercise 4.1.3

Three metal cubes whose edges measure 3cm, 4cm and 5cm respectively are melted to form a single cube. Find (i) Side length (ii) total surface area of the new cube. What is the difference between the total surface area of the new cube and the sum of total surface areas of the original three cubes?

Solution:

Three metal cubes whose edges measure 3cm, 4cm and 5cm respectively are melted to form a single cube.

A closed figure having four sides formed by joining four points, no three of which are collinear, in an order is called quadrilateral.

For example:

Quadrilateral is the union of four line segments that join four coplanar points; no three of which are collinear and each segment meet exactly two other lines, each at their end point.

A quadrilateral is convex if each of the internal angle is less than 180˚. Otherwise it is concave quadrilateral.

Properties of quadrilateral – Quadrilaterals

Let ABCD be a quadrilateral.

Then points A, B, C and D are called the vertices of the quadrilateral.

The segments AB, BC, CD and DA are the four sides of the quadrilateral.

The angles ∠DAB, ∠ABC, ∠BCD, and ∠CDA are the four angles of the quadrilateral.

The segments AC and BD are called as the diagonals of the quadrilateral.

Adjecent sides and opposite sides:

The sides of a quadrilateral are said to be adjacent sides or consecutive sides. If they have a common end point. In the adjoining figure, AB and AD are adjacent or consecutive sides. Identify the other pair of adjacent sides.

Two sides of a quadrilateral are said to be opposite angles, if they do not contain a common side. Here ∠DAB and ∠BCD are opposite angles. Identify other pair of opposite angles.

Adjacent angles and opposite angles:

The two angles of a quadrilateral are adjacent angles or consecutive angles. If they have a common side to them. Thus ∠DAB and ∠ABC are adjacent angles

Two angles of a quadrilateral are said to be opposite angles, if they do not contain a common side. Here ∠DAB and ∠BCD are opposite angles. Identify other pair of opposite angles.

Diagonal Property:

The diagonal AC divides the quadrilateral into two triangles, namely, triangles ABC and triangles ADC. Name the triangles formed when the diagonal BD is drawn.

Rhombus:

A rhombus is a parallelogram in which all the four sides are equal.

Example 10: The diagonals of a rhombus are 24cm and 10cm. Calculate the area of the rhombus.

Solution:

We are given that AC = 24cm; BD = 10cm. We know that the diagonals of a rhombus bisect each other at right angles. Let O be the point of intersection of these diagonals. Then we have AO = CO = 12cm and BO = DO = 5cm. We also know that AOD is right angled triangle. Hence the area of triangle AOD is

1/2 x OA x OD = 1/2 x 12 x 5 = 30 cm2

Since a rhombus has four congruent right triangles, is area is 4 x 30 = 120 cm2

Square

A square is a parallelogram, in which

(i) in which all the sides are equal

(ii) each angle is right angle

(iii) diagonals are equal

(iv) diagonals bisect at right angles.

Example 12: A field is in the shape of a sqaure with side 20m. A pathway of 2m width is surrounding it. Find the outer perimeter of the pathway.

Solution:

Width of the pathway is 2m. Length of the side of the outer sqaure = (20 + 2 + 2) = 24m

Hence, perimeter 4 x 24 = 96m

Kite:

Kite is a quadrilateral in which two of isosceles triangles are joined along the common base.

Example 14: In the figure PQRS is a kite; PQ = 3cm and QR = 6cm. Find the perimeter of PQRS.

Solution:

We have PQ = PS = 3cm, QR = SR = 6cm.

Hence, the perimeter = PQ + QR + RS + PS

3 + 6 + 6 + 3 = 18cm

Quadrilaterals – EXERCISE 3.5.5

1. The sides of the rectangle are in the ratio 2:1. The perimeter is 30 cm. Calculate the measure of all the sides.Solution:

10. Let ABCD be a rhombus and ∟ABC = 124°. Calculate ∟A, ∟D and ∟C. In a rhombus, opposite angles are equal

Solution:∠B = ∠D = 124°∠A + ∠B =180°Consecutive angles

∠A + 124° = 180°

∠A + 124° = 180°∠A = 180-124=56°∴ ∠A = 56° ; ∠B = 124° and ∠C = 56°

11. Rhombus is a parallelogram: justify.

Solution:Rhombus has all the properties of parallelogram i.ea) Opposite sides are equal and parallel.b) Opposite angles are equal.c) Diagonals bisect each other∴ Rhombus is a parallelogram.

12. In a given square ABCD, if the area of triangle ABD is 36 cm2 , find (i) the area of triangle BCD; (ii) the area of the square ABCD.

Solution:

ABCD is a squareBd is the diagonalThe diagonal divides the square into two congruent triangles∴ Δ ABD ≅ Δ BCD D C∴ Area of Δ ABD =area of Δ BCDArea of Δ ABD= 36 cm² (given)∴ Area of Δ BCD = 36cm²Area of the square ABCD = Area of Δ ABD + Area of ΔBCD=36cm² +36cm²Area of the squre ABCD =72cm²

13. The side of a square ABCD is 5 cm and another square PQRS has perimeter equal to 40cm. Find the ratio of the area ABCD to the area of PQRS.

14. A square field has side 20m. Find the length of the wire required to fence it four times.

Solution:Length of one side of the square = 20 RsLength of wire required to fence one round = 4×20Length of wire required to fence four rounds = 80m= 4×80m= 320m

15. List out the differences between square and rhombus.

Solution:Square1. All the angles are equal2. Diagonals are equal3. Area = side × side=(s)²

Rhombus

Opposite angles are equal

2. Diagonals are unequal

3. Area= 1/2 × Product of diagonals = 1/2 x d­1 d2

Example 16. Four congruent rectangles are placed as shown in the figure. Area of the outer square is 4 times that of the inner square. What is the ratio os length to breadth of the congruent rectangles?

Solution:

Let the length of rectangles be ‘a’ and breadth be ‘b’Side of outer square –(a+b) unitsSide of outer square (a-b) unitsArea of outer square = 4 times area of inner squareArea of ABCD = 4(Area of PQRS)(a + b)² = 4(a-b)